Easier way to guarantee integer input through Scanner?

Question

For a program I am writing, I need to ask a user for an integer between 1 and 8. I've tried multiple (cleaner) ways of doing this but none of them worked, so I'm left with this:

    int x = 0;
    while (x < 1 || x > 8)
    {   
        System.out.print("Please enter integer  (1-8): ");

        try
        {
            x = Integer.parseInt(inputScanner.next());
        }
        catch(NumberFormatException e)
        {
            x = 0;
        }
    }

Where inputScanner is a Scanner. Surely there is a better way?

Solution 1

Scanner does regular expressions, right? Why not check if it matches "^[1-8]$" first?

Solution 2

Using the nextInt() is already an improvement compare to simply using the next() method. And before that, you can use the hasNextInt() to avoid haing all this bunch of useless exceptions.

Resulting in something like this:

int x = 0;
do {
  System.out.print("Please...");
  if(scanner.hasNextInt()) x = scanner.nextInt();
  else scanner.next();
} while (x < 1 || x > 8);

Solution 3

I had to do a graphic interface calculator (works only with Integers), and the problem was, that the Tests didn't allow any Exceptions to be thrown if the input wasn't Integer. So I couldn't use

try { int x = Integer.parseInt(input)} catch (Exception e) {dosomethingelse}

Because Java programs generally treat an input to a JTextField as a String I used this:

if (input.matches("[1-9][0-9]*"){ // String.matches() returns boolean
   goodforyou
} else {
   dosomethingelse
}

// this checks if the input's (type String) character sequence matches
// the given parameter. The [1-9] means that the first char is a Digit
// between 1 and 9 (because the input should be an Integer can't be 0)
// the * after [0-9] means that after the first char there can be 0 - infinity
// characters between digits 0-9

hope this helps :)

Solution 4

You could try something like this:

Scanner cin = new Scanner(System.in);
int s = 0;    
boolean v = false;
while(!v){
    System.out.print("Input an integer >= 1: ");

    try {    
        s = cin.nextInt();
        if(s >= 1) v = true;
        else System.out.println("Please input an integer value >= 1.");
    } 
    catch(InputMismatchException e) {
        System.out.println("Caught: InputMismatchException -- Please input an integer value >= 1. ");
        cin.next();
    }
}

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Comments

[Logan Serman]

That results in an infinite loop if you provide a string.

[WolfmanDragon]

While I know your code is syntactically correct, having a if then on one line without braces is none standard. From Sun: Note: if statements always use braces {}. Avoid the following error-prone form: if (condition) statement; //AVOID! THIS OMITS THE BRACES {}!

[palantus]

And he also omitted the semicolon after the while(), yet I somehow managed to understand his meaning. :)

[Joe Phillips]

I thought a snip-it was one of those things the kids put around their ankle and jumped over while it spun in circles?

[OscarRyz]

@Gizmo: I've heard that same reasoning thousands of times in my life. A lot of bugs have been introduced by that same reason. Although it is correct that such a small piece of code shouldn't be taken so serious, it ultimately reflects your practices. Is a matter of good habits.

[OscarRyz]

Does it worth to bring a new lib just for one function? : - /

[WolfmanDragon]

This site is about learning, it should not be about ego. When people point out errors that I make here, and I do make errors, I admit my mistakes and go on with life. It is much harder to break a bad habit than to a create a good habit from the start. Remember the original slogan for this site? ;)

[gizmo]

@Oscar > Ok, mister the medium. So, reading my snippet, did you deduce that, IRL, all my code is passed through a formatter at saving time? That I'm working using the TDD/BDD methodologies? That I introduced and guaranteed the good practices you're talking about in my company? I don't think so.